The burial cloth of an Egyptian mummy is estimated to contain 56% of the carbon-14 it contained originally. How long ago was the mummy buried? (the half-life of carbon-14 is 5730). Please round the answer to the nearest tenth. I have figured that:
m(t) = Moe(-ln2/5730) is a starting point
Let
P14 = fraction of C14 found in sample (0.56)
T1/2 = radioactive half life (5730 years)
Then
Age of sample
=T1/2*(ln(P14)/ln(0.5)) (Before present)
=5730*(ln(0.56)/ln(0.5)) (BP)
=4793 (BP)
To solve this problem, we can use the decay equation for carbon-14:
m(t) = M0 * e^(-ln2/5730 * t)
Given that the burial cloth of the Egyptian mummy contains 56% of the carbon-14 it originally had, we can write this as:
0.56 = e^(-ln2/5730 * t)
To determine how long ago the mummy was buried, we need to solve for t. Start by taking the natural logarithm of both sides:
ln(0.56) = -ln2/5730 * t
Now, divide both sides by -ln2/5730:
t = ln(0.56) / (-ln2/5730)
Using a calculator, we can calculate t to be approximately 7224.5 years.
Therefore, the mummy was buried approximately 7224.5 years ago.
To find out how long ago the mummy was buried, we can use the formula for exponential decay with the half-life of carbon-14. The formula can be written as:
m(t) = Mo * e^(-ln2/5730 * t)
Where:
m(t) is the amount of carbon-14 remaining after time t
Mo is the initial amount of carbon-14
ln2 is the natural logarithm of 2
Given that the burial cloth of the mummy contains 56% of its original carbon-14, we can say:
0.56Mo = Mo * e^(-ln2/5730 * t)
Now, to solve for t, we can cancel out the initial amount Mo from both sides:
0.56 = e^(-ln2/5730 * t)
To isolate t, we can take the natural log of both sides:
ln(0.56) = -ln2/5730 * t
Next, divide both sides by -ln2/5730 to solve for t:
t = ln(0.56) / (-ln2/5730)
Using a calculator, we can determine the value of t:
t ≈ 3.835 * 5730
t ≈ 22023.405
Rounded to the nearest tenth, the mummy was buried approximately 22,023.4 years ago.